A352522 Triangle read by rows where T(n,k) is the number of integer compositions of n with k weak nonexcedances (parts on or below the diagonal).
1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 2, 3, 4, 3, 3, 1, 3, 4, 8, 6, 6, 4, 1, 4, 7, 12, 13, 12, 10, 5, 1, 5, 13, 16, 26, 24, 22, 15, 6, 1, 7, 19, 27, 43, 48, 46, 37, 21, 7, 1, 10, 26, 47, 68, 90, 93, 83, 58, 28, 8, 1, 14, 36, 77, 109, 159, 180, 176, 141
Offset: 0
Examples
Triangle begins: 1 0 1 1 0 1 1 1 1 1 1 3 1 2 1 2 3 4 3 3 1 3 4 8 6 6 4 1 4 7 12 13 12 10 5 1 5 13 16 26 24 22 15 6 1 7 19 27 43 48 46 37 21 7 1 10 26 47 68 90 93 83 58 28 8 1 For example, row n = 6 counts the following compositions: (6) (15) (114) (123) (1113) (11112) (111111) (24) (42) (132) (1311) (1122) (11121) (33) (51) (141) (2112) (1131) (11211) (231) (213) (2121) (1212) (12111) (222) (2211) (1221) (312) (3111) (21111) (321) (411)
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)
- MathOverflow, Why 'excedances' of permutations? [closed].
Crossrefs
Programs
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Mathematica
pw[y_]:=Length[Select[Range[Length[y]],#>=y[[#]]&]]; Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],pw[#]==k&]],{n,0,15},{k,0,n}]
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PARI
T(n)={my(v=vector(n+1, i, i==1), r=v); for(k=1, n, v=vector(#v, j, sum(i=1, j-1, if(k>=i,x,1)*v[j-i])); r+=v); [Vecrev(p) | p<-r]} { my(A=T(10)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Jan 19 2023